C. Team

Problem - C - Codeforces

Now it's time of Olympiads. Vanya and Egor decided to make his own team to take part in a programming Olympiad. They've been best friends ever since primary school and hopefully, that can somehow help them in teamwork.

For each team Olympiad, Vanya takes his play cards with numbers. He takes only the cards containing numbers 1 and 0. The boys are very superstitious. They think that they can do well at the Olympiad if they begin with laying all the cards in a row so that:

• there wouldn't be a pair of any side-adjacent cards with zeroes in a row;
• there wouldn't be a group of three consecutive cards containing numbers one.

Today Vanya brought n cards with zeroes and m cards with numbers one. The number of cards was so much that the friends do not know how to put all those cards in the described way. Help them find the required arrangement of the cards or else tell the guys that it is impossible to arrange cards in such a way.

Input

The first line contains two integers: n (1 ≤ n ≤ 106) — the number of cards containing number 0; m (1 ≤ m ≤ 106) — the number of cards containing number 1.

Output

In a single line print the required sequence of zeroes and ones without any spaces. If such sequence is impossible to obtain, print -1.

Examples

input

Copy

1 2

output

Copy

101

input

Copy

4 8

output

Copy

110110110101

input

Copy

4 10

output

Copy

11011011011011

input

Copy

1 5

output

Copy

-1
题意：给一组只有0和1的数组，排序满足条件：

1.两个相邻的零不能在一起

2.三个1不能在一起

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
// ctrl+shift+C 注释
//ctrl+shift+x 取消
#define int long long
#define YES cout<<"YES"<
#define NO cout<<"NO"<
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;
typedef pair<int,int> PII;
const int N=2e5+10;
const ll M=1e18+10;
const int mod=1e9+7;
int a[N],sum[N];
priority_queue<int,vector<int>,greater<int> >pq;
set<int>se;
map<int,int>mp;
queue<int>qu;
vector<int>v;
deque<int>de;
int n,m;
void solve()
{
    cin>>n>>m;
    if(m>n*2+2||n>m+1)
    {
        cout<<-1<
        return;
    }
    int f=0;
    if(m==n)// 010101
    {
        for(int i=0;i
        {
            cout<<"10";
        }
    }
    else if(m-n<=2&&m>n)// n=2 m=3 10101
    {
        for(int i=1;i<=n;i++)
        {
            cout<<"10";
        }
        for(int i=1;i<=m-n;i++)
        cout<<"1";
    }
    else if(n-1==m)// n=3 m=2  01010
    {
        for(int i=1;i<=m;i++)
        {
            cout<<"01";
        }
        cout<<0;
    }
    else if(m2)// 0多 n=4 m=7  11011011010
    {
        int res=n*2-m;
        for(int i=1;i<=n-res;i++)
        {
           cout<<"110";
        }
        for(int i=1;i<=res;i++)
        {
            cout<<"10";
        }
    }
   else if(m==n*2)
    {
        for(int i=0;i
        {
            cout<<"110";
        }
    }
    else if(m>n*2&&m<=n*2+2)//1多 n=4 m=9  1101101101101
    {
         for(int i=0;i
         {
             cout<<"110";
         }
         int res=m-n*2;
         for(int i=1;i<=res;i++)
         {
             cout<<1;
         }
    }
    cout<
}
signed main()
{
   int t=1;
   //cin>>t;
   while(t--)
   {
       solve();
   }
}